Upload
duongnhu
View
220
Download
0
Embed Size (px)
Citation preview
ESE 271 / Spring 2013 / Lecture 6
Last time: Proportionality and Superposition.Example we considered last time:Example we considered last time:
For linear circuits
For any
Proportionality: Scaling of all independent sources
Superposition:Response of circuit to several sources
1
Scaling of all independent sources scales all circuit variables.
is equal to sum of responses to each individual source.
ESE 271 / Spring 2013 / Lecture 6
Last time: Nodal Analysis.
Idea:If knownIf known
Then:
Reference node.All voltages will be calculated with respect to this reference.
2
ESE 271 / Spring 2013 / Lecture 6
Last time: Example of Nodal AnalysisNode 2
Node 1Node 3
Let’s choose Node 4 as a reference node hence V = 0
Node 4
reference node, hence V4 = 0.
3
ESE 271 / Spring 2013 / Lecture 6
Last time: Nodal Analysis ‐ Equation patternNode 2
Node 1Node 3
Let’s choose Node 4 as a reference node hence V = 0
Node 4
reference node, hence V4 = 0.
Currents from current h h
Sum of conductances of the elements connected
Other nodal voltages multiplied by MINUS conductances of the
4
sources that enter the node
the elements connected to node multiplied by
nodal voltage
by MINUS conductances of the elements connecting other nodes to the equation node.
ESE 271 / Spring 2013 / Lecture 6
Last time: Nodal analysis of circuits containing voltage sources
Can not find current even if nodal voltages are known. What to do?
Example we have considered last time
Supernode A contains Node 1 and Reference Node, hence:
Supernode B contains Node 4 andSupernode B contains Node 4 and Node 5, hence:
Summary: In this circuit there are five non‐reference nodes and two voltage sources, hence only three unknown nodal voltages:
5
nodal voltages:
ESE 271 / Spring 2013 / Lecture 6
Example 2
6
ESE 271 / Spring 2013 / Lecture 6
Example 3 – with dependent current source
Reference node
Unknown nodal voltages
7
ESE 271 / Spring 2013 / Lecture 6
Example 4 – with dependent voltage source
Reference node
Unknown nodal voltages:
But hence
Only are unknown
Also
KCL at node 4:
KCL at supernode:
8
ESE 271 / Spring 2013 / Lecture 6
Mesh analysisIn its basic form it is suitable for planar circuits onlyIn its basic form it is suitable for planar circuits only.
Hence not as universal as nodal analysis but more convenient in certain cases.
EXAMPLE – circuit containing four meshes (loops with no loops inside)
ACTION ITEMS1. Assign current to each mesh.2. Write KVL for each mesh.3. Solve system of linear equations for mesh currents.4. Find actual currents and voltages.
Mesh 1:
Mesh 2:
Mesh 3:
9
Mesh 4:
ESE 271 / Spring 2013 / Lecture 6
Mesh analysis: Example 1.
1. Assign current to each mesh.2. Write KVL for each mesh.
3. Solve for mesh currents:
4. Find actual currents and voltages.
10
ESE 271 / Spring 2013 / Lecture 6
Mesh analysis: Example 2.
11
ESE 271 / Spring 2013 / Lecture 6
Mesh analysis of circuits containing current sources.Voltage drop across current source is not definedVoltage drop across current source is not defined.
However, presence of current source reduces number of equations by one.
EXAMPLE – circuit containing four meshes and two current sources
Four meshes, hence, four mesh currents:
BUT:
Only two mesh currents remain unknown.Hence, only two equations are needed.Write KVL for supermeshes.
12
f p
ESE 271 / Spring 2013 / Lecture 6
Example 1.
Four meshes, hence, four mesh currents:
BUT:
Hence only two unknowns: and
Observe:
KVL for mesh 2:
KVL for supermesh 1+3+4:
13
ESE 271 / Spring 2013 / Lecture 6
Example 1 ‐ cont.
14
ESE 271 / Spring 2013 / Lecture 6
Example 2.
Two mesh currents:
BUT:
15
ESE 271 / Spring 2013 / Lecture 6
Nodal analysis of circuits with OpAmps.Example 1Example 1.
i t l h t
KCL at Node 2:
virtual short
No current into OpAmp
KCL at Node 3:
16
ESE 271 / Spring 2013 / Lecture 6
Nodal analysis of circuits with OpAmps.Example 2Example 2.
17
ESE 271 / Spring 2013 / Lecture 6
Example 3.
18